Problem: Consider a new derivative that at time $T$ pays $Y =(S(T_0) − S(T))^+$

where $0 < T_0 < T$ is a fixed date.

(i) Show that the arbitrage-free of Y at time $t = T_0$ is given by $\pi_{T_0} (Y ) =pS(T_0)$ where p is independent of the stock price.

(ii) Determine the arbitrage-free price of derivative $Y$ at time $t < T_0$.

To value the contract I would use the Black-Scholes PDE solution $E^{Q}((S(T_0) − S(T))^+|\mathscr{F}_{T_{0}})=E^{Q}((S(T_0) − S(T)) \:1_{S(T_0\geqslant S(T))}|\mathscr{F}_{T_{0}})$

I know that $S(T_0)-S(T)$ is independent of $\mathscr{F}_{T_0}$. But it is not independent of $1_{S(T_0\geqslant S(T))}$, so I cannot split them into two expected values. Even if I did, I would not get the result. I tried to solve using the exponential form of S(T) but got nowhere.

I though that when $t=T_0$ then $Y$ is a put option which implies that $E^{Q}((S(T_0) − S(T))^+|\mathscr{F}_{T_{0}})=S(T_0)(e^{T-T_0}\Phi(-d_2)-\Phi(-d_1))$ where d_2 and d_1 do not depend on the stock.


Is my solution correct?

How do I solve the second question when $t<T_0$? I think I cannot used a pre-defined result.

Thanks in advance!


1 Answer 1


At time $T_0$, the strike price becomes known and the option turns into a ``normal'' put option, i.e. \begin{align*} V(T_0,S_{T_0}) &= S_{T_0}e^{-r(T-T_0)}\Phi(-d_2)-S_{T_0}e^{-q(T-T_0)}\Phi(-d_1) \\ &= S_{T_0}\underbrace{\left(e^{-r(T-T_0)}\Phi(-d_2)-e^{-q(T-T_0)}\Phi(-d_1)\right),}_{=:p} \end{align*} where $p$ is indeed independent of the stock price because \begin{align*} d_{1,2}=\frac{r-q\pm\frac{1}{2}\sigma^2}{\sigma}\sqrt{T-T_0}. \end{align*}

In general, for $t> T_0$, the Black-Scholes formula is \begin{align*} V(t,S_t) &= S_{T_0}e^{-r(T-t)}\Phi(-d_2)-S_te^{-q(T-t)}\Phi(-d_1), \end{align*} where \begin{align*} d_{1,2}=\frac{\ln\left(\frac{S_t}{S_{T_0}}\right)+\left(r-q\pm\frac{1}{2}\sigma^2\right)(T-t)}{\sigma\sqrt{T-t}}. \end{align*} Of course, $\lim\limits_{t\downarrow T_0}V(t,S_t)=S_{T_0}p$.

For $t<T_0$, the option value is \begin{align*} V(t,S_t)&=e^{-r(T-t)}\mathbb{E}^\mathbb{Q}_t\left[\max\{S_{T_0}-S_T,0\}\right] \\ &=e^{-r(T-t)}\mathbb{E}^\mathbb{Q}_t\left[\mathbb{E}^\mathbb{Q}_{T_0}\left[\max\{S_{T_0}-S_T,0\}\right]\right] \\ &=e^{-r(T_0-t)}\mathbb{E}^\mathbb{Q}_t\left[e^{-r(T-T_0)}\mathbb{E}^\mathbb{Q}_{T_0}\left[\max\{S_{T_0}-S_T,0\}\right]\right]\\ &=e^{-r(T_0-t)}\mathbb{E}^\mathbb{Q}_t\left[S_{T_0}p\right] \\ &=pe^{-r(T_0-t)}S_te^{(r-q)(T_0-t)} \\ &=S_te^{-q(T_0-t)}\left(e^{-r(T-T_0)}\Phi(-d_2)-e^{-q(T-T_0)}\Phi(-d_1)\right). \end{align*} Of course, $\lim\limits_{t\uparrow T_0}V(t,S_t)=S_{T_0}p$.

  • 1
    $\begingroup$ In case of interest, this option is also pricable analytically under Heston stochastic vol, eg. core.ac.uk/download/pdf/34226925.pdf $\endgroup$
    – StackG
    Commented Feb 20, 2021 at 23:11
  • 1
    $\begingroup$ As @StackG correctly adds, the option is pricable in closed form for many models with closed-form solution for European put options. To keep the calculations simple, we'd need $p$ to be independent of $S_{T_0}$. This applies to exponential Lévy processes (e.g. Merton and Kou's jump diffusion models as well as NIG, VG and CGMY) and stochastic volatility models (e.g. Heston or Grasselli's 4/2 model). $\endgroup$
    – Kevin
    Commented Feb 20, 2021 at 23:36

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